To solve for [tex]\( y \)[/tex] in the equation
[tex]\[
\left|\begin{array}{ccc}
y^2 & y & 1 \\
8 & 4 & 10 \\
9 & 3 & 6
\end{array}\right| = 60
\][/tex]
we need to follow these steps:
1. Set Up the Determinant Equation:
First, we write out the determinant of the given 3x3 matrix.
2. Calculate the Determinant:
The determinant of a 3x3 matrix
[tex]\[
\begin{vmatrix}
a_{11} & a_{12} & a_{13} \\
a_{21} & a_{22} & a_{23} \\
a_{31} & a_{32} & a_{33}
\end{vmatrix}
\][/tex]
is calculated as:
[tex]\[
a_{11}(a_{22}a_{33} - a_{23}a_{32}) - a_{12}(a_{21}a_{33} - a_{23}a_{31}) + a_{13}(a_{21}a_{32} - a_{22}a_{31})
\][/tex]
3. Substitute Values from the Matrix:
For our specific matrix, we substitute in the values:
[tex]\[
\left|\begin{array}{ccc}
y^2 & y & 1 \\
8 & 4 & 10 \\
9 & 3 & 6
\end{array}\right|
\][/tex]
So, using the formula:
[tex]\[
y^2 \left(4 \cdot 6 - 10 \cdot 3\right) - y\left(8 \cdot 6 - 10 \cdot 9\right) + 1 \left(8 \cdot 3 - 4 \cdot 9\right)
\][/tex]
4. Simplify the Determinant Expression:
[tex]\[
y^2 (24 - 30) - y (48 - 90) + 1 (24 - 36)
\][/tex]
[tex]\[
y^2 (-6) - y (-42) + (-12)
\][/tex]
[tex]\[
-6y^2 + 42y - 12
\][/tex]
5. Set the Determinant Equal to 60:
The determinant is given to be equal to 60, so we set up the equation:
[tex]\[
-6y^2 + 42y - 12 = 60
\][/tex]
6. Rearrange and Solve the Equation:
Move 60 to the left side to set the equation to 0:
[tex]\[
-6y^2 + 42y - 12 - 60 = 0
\][/tex]
[tex]\[
-6y^2 + 42y - 72 = 0
\][/tex]
Divide by -6 to simplify:
[tex]\[
y^2 - 7y + 12 = 0
\][/tex]
7. Factor the Quadratic Equation:
Factor the quadratic equation:
[tex]\[
(y - 3)(y - 4) = 0
\][/tex]
8. Solve for [tex]\( y \)[/tex]:
Set each factor to zero and solve:
[tex]\[
y - 3 = 0 \quad \Rightarrow \quad y = 3
\][/tex]
[tex]\[
y - 4 = 0 \quad \Rightarrow \quad y = 4
\][/tex]
So, the solutions for [tex]\( y \)[/tex] are [tex]\( y = 3 \)[/tex] and [tex]\( y = 4 \)[/tex].
Thus, the correct answer is:
[tex]\[
\boxed{3, 4}
\][/tex]
